penalized LOD scores with stepwiseqtl(additive.only=TRUE).

[JC]

Karl,

I am trying to run stepwiseqtl() with the additive.only=TRUE argument specified (as in your previous post). I am also specifying the output of scanone with 1000 permutations as the value for the penalties argument. I am working with a phenotype with one major QTL which shows up prominently in the scanone output. However when I run stepwiseqtl() two unusual things happen:

1) I get a warning indicating that "In stepwiseqtl(pop11, pheno.col = 43, qtl = rqtl, method = "hk",  :

  penalties should have length 3" This makes me believe it's still looking for the output of calc.penalties() from a 2D genome scan even though I specified additive.only=TRUE in the stepwiseqtl() function. 

2) This may be an artifact of the above, but after running stepwiseqtl() the model that wins out is the null model with no QTL. Is it unusual for a QTL to be detected by scanone() and then removed by stepwiseqtl()? I've seen this QTL before and I know it's real, so I'm a little leary of the stepwiseqtl() results. 

Do you have any thoughts? Is there a way for me to format the output from scanone() to be more appropriate for stepwiseqtl()? calc.penalties() will only accept the output from scantwo().

Best Regards,

Josh

[Karl Broman]

1) It wants a threshold as a penalty. Use summary() with the output from scanone permutations to get a threshold, and give that as the penalty.

2) If scanone gives you a QTL with a LOD score that is above the penalty from (1), then it will be favored over the null model.

karl

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